Answer:
Choice D
Step-by-step explanation:
The function will be continuous everywhere except where the function in the denominator is 0. We, therefore determine the value(s) of x for which the function will be 0;
[tex]x^{2} -6x+8=0\\x^{2} -2x-4x+8=0\\x(x-2)-4(x-2)=0\\(x-4)(x-2)=0\\x=4\\x=2[/tex]
Therefore, the points x= 2 and x =4 are points of discontinuity
